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A Special Functional Identity in Prime RingsTsai-Lien Wong a ba Department of Applied Mathematics , National Sun Yat-Sen University , Kaohsiung ,Taiwan , P.R. Chinab Department of Applied Mathematics , National Sun Yat-Sen University , Kaohsiung ,Taiwan , 804 , P.R. ChinaPublished online: 10 Oct 2011.

To cite this article: Tsai-Lien Wong (2004) A Special Functional Identity in Prime Rings, Communications in Algebra, 32:1,363-377, DOI: 10.1081/AGB-120027872

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A Special Functional Identity in Prime Rings#

Tsai-Lien Wong*

Department of Applied Mathematics, National Sun Yat-Sen University,Kaohsiung, Taiwan, P.R. China

ABSTRACT

Let A be a prime ring with char(A) 6¼ 2, let f :A!A be an additive map and let dbe a nonzero derivation of A. If d([ f(x), x])¼ 0 for all x2A, then [ f(x), x]¼ 0 for

all x2A.

Key Words: Functional identity; Prime ring; Additive mapping; Commutingmapping; Derivation; Centralizer.

1. INTRODUCTION

A map f from a ring A into itself is said to be commuting if [f(x), x]¼ 0 for allx2A. The study of such maps has the origin in the result of Posner (1957, Theorem 2)from 1957, which states that there are no nonzero commuting derivations on non-commutative prime rings. Bresar (1993, Theorem 3.2) proved that every additivecommuting map f on a prime ring A is of the form f(x)¼ lxþ m(x) where l is anelement in the extended centroid C of A and m :A!C is an additive map. This resulthas been extended in various directions and in fact it initiated the theory of

#Communicated by R. Wisbauer.

*Correspondence: Tsai-Lien Wong, Department of Applied Mathematics, National SunYat-Sen University, Kaohsiung, Taiwan 804, P.R. China; E-mail: tlwong@math.nsysu.edu.tw.

COMMUNICATIONS IN ALGEBRA�

Vol. 32, No. 1, pp. 363–377, 2004

363

DOI: 10.1081/AGB-120027872 0092-7872 (Print); 1532-4125 (Online)

Copyright # 2004 by Marcel Dekker, Inc. www.dekker.com

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functional identities. We refer the reader to Bresar (2000) for an introductoryaccount on functional identities and their applications.

In Bresar (1993, Proposition 3.1), it was also proved that if char(A) 6¼ 2 and anadditive map f :A!A is not commuting, then it is also not centralizing, that is tosay, the set consisting of elements of the form [ f(x), x], x2A, cannot be containedin the center of A. This suggests the question: how large is this set? In the specialcases when f is a derivation or an automorphism this question has been extensivelystudied (see Bresar and Vukman, 1993; Chebotar, 1996; Chebotar and Lee, 2001;Chebotar et al., 2002; Chuang and Lee, 2002; Lee and Wong, 2002; Mayne, 1976,1992; Wong, 2002). Our goal is to show that for any additive map, not only thatthis set is not contained in the center, but it has the trivial centralizer; even moregenerally, it cannot be contained in the kernel of a nonzero derivation. So, our mainresult is

Theorem 1. Let A be a prime ring with char(A) 6¼ 2, let f : A!A be an additivemap and let d be a nonzero derivation of A. If d([ f(x), x])¼ 0 for all x2A, thenf is commuting (and hence there exist l2C and an additive map m : A!C such thatf(x)¼ lxþ m(x) for all x2A).

We remark that the case when char(A)¼ 2 must indeed be excluded. Namely,given a nonzero element a in a prime ring A with char(A)¼ 2 such that a2¼ 0, wethen have [a, [a, x]]¼ 0, x2A. Therefore, if a map f is given by f(x)¼ n(x)a, wheren is any nonzero additive map with the range in the center of A, and d is the innerderivation induced by a, then d([ f(x), x])¼ 0 for all x2A, but f is not commuting(namely, a is not central since a2¼ 0). We remark that this example is a modificationof the one given in Bresar and Vukman (1993, p. 134).

Theorem 1 somehow indicates a new possible way how to generalize existingresults on functional identities. Since functional identities have proved to be applic-able to various areas, we hope that this new way shall also turn out to be useful. Con-cerning Theorem 1, the least we can say that it covers several apparently unrelatedresults from the literature. First of all, in the special case when f is a derivation, The-orem 1 essentially coincides with the result of Bresar and Vukman (1993, Lemma 1).Further, if f is a generalized inner derivation, i.e., f(x)¼ ax� xb for some fixed ele-ments a and b, then [ f(x), x]¼da(x)x� xdb(x) where da and db are correspondinginner derivations, and so Theorem 1 yields an extension of (a special case of) Bresar(1993, Theorem 4.1). In the case when f¼� is an involution and d is an inner deriva-tion, we get the following corollary: If a prime ring A with involution does not satisfyS4 (so that � is not a commuting map, Herstein, 1976, Theorem 2.5.2) andchar(A) 6¼ 2, then only the elements from the center commute with every elementof the form [x�, x], x2A. Note that elements of the form [x�, x] are just special exam-ples of symmetric elements, so this assertion can be viewed (at least whenchar(A) 6¼ 2) as a generalization of a well-known Herstein’s result (1969, Theorem1.7) stating that an element must be in the center if it commutes with every symmetricelement. Finally, suppose f is of the form f(x)¼ n(x)a where n is a central map and ais a fixed element (i.e. f is a kind of a ‘‘rank one’’ map). Then Theorem 1 gives a ver-sion of Posner’s theorem on the product of derivations (Posner, 1957, Theorem 1)(thus, the case when f is of ‘‘finite rank’’ greater than one might be of some interest).

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From the technical point of view, the present article was strongly influenced bythe recent work of Beidar et al. (2002) where, as an application of the generaltheory of functional identities with r-independent coefficients developed there, thefunctional identity [ f(x), d(x)]¼ 0 is studied (here d is a derivation and f is an addi-tive map). The first parts of the proof of Theorem 1 are just modifications of those ofBeidar et al. (2002, Theorem 4.1).

As already mentioned, there are many results in the literature concerning the sizeof the set of all [ f(x), x], x2A (or all x in some appropriate subset of A), where f aderivation or an automorphism. A typical result states that the subring (or some-times just an additive subgroup) generated by this set contains a nontrivial (two-sided, Lie or one-sided) ideal of A. The following example shows that in generalwe cannot get as general results for any additive map f.

Example 2. Let B be any noncommutative ring and let A¼B[X ] be the ring of poly-nomials with coefficients from B. Define an additive map f :A!A by f(bX2n)¼ 0and f(bX2nþ1)¼ bX2n for every b2B and n� 0. Since B is noncommutative, f isnot commuting. Note that [f(bXi), b0Xj]þ [f(b0Xj), bXi] can be nonzero only whenone of i, j is even and the other one is odd, in which case this expression is equalto ±[b, b0]Xiþj�12B[X2]. Whence we see that [ f(x), x]2B[X2] for every x2A, andso the subring A0 generated by all [ f(x), x], x2A, is also contained in B[X2]. However,B[X2] contains neither nonzero one sided ideals nor noncentral Lie ideals of A, and sothe same is true for A0.

2. PROOF

In what follows we assume that the conditions of Theorem 1 are fulfilled, i.e.,A is a prime ring with char(A) 6¼ 2, d is a nonzero derivation of A and f :A!A isan additive map such that

dð½fðxÞ; x�Þ ¼ 0 for all x 2 A; ð1Þ

and hence also

dð½fðxÞ; y�Þ þ dð½fðyÞ; x�Þ ¼ 0 for all x; y 2 A: ð2Þ

These two identities will be often used in the sequel without explicit mention. By Cwe denote the extended centroid of A, by B¼ACþC the central closure of A, by Qs

the symmetric Martindale ring of quotients of A and by Q either the maximal left orthe maximal right ring of quotients of A. For definitions and basic properties ofthese, as well as of some other concepts appearing in the present article we referthe reader to Beidar et al. (1996). Let us remark that with no loss of generality wemay assume that d is a derivation defined on the whole Q (Beidar et al., 1996,Proposition 2.5.1).

Our goal is to show that [ f(x), x]¼ 0 for all x2A. We first note that it suffices toshow that [f(y), y]¼ 0 for all y in some nonzero ideal I of A. Indeed, in this case byBresar (1994, Corollary 4.2) there exists l2C such that f(y)� ly2C for all y2 I. Fixx2A. By (2) we have d([ f(x)� lx, y])¼ 0 for any y2 I. That is, the composition of

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two derivations, y 7! [ f(x)� lx, y] and d, is equal to zero on the prime ring I. By thewell-known Posner’s result (1957, Theorem 1) it follows that one of these two deriva-tions is zero (to be precise, we cannot directly apply Posner’s theorem since these twoderivations do not necessarily map I into itself, but fortunately the same proof worksin this slightly more general situation). Since d 6¼ 0 it follows that f(x)� lx2C andso [ f(x), x]¼ 0, as desired.

The proof of our goal is broken up into the series of steps. As in the proof ofBeidar et al. (2002, Theorem 4.1) we first use the result of Beidar et al. (1999) toreduce the problem to the case when d is an X-inner derivation.

2.1. Reduction to the X-Inner Case

Throughout this subsection we assume that d is an X-outer derivation. Let usfirst consider the case when A is not a GPI ring. Note that (2) can be written as

fðxÞdðyÞ þ fðyÞdðxÞ þ dðfðxÞÞyþ dðfðyÞÞx� dðyÞfðxÞ � dðxÞfðyÞ� ydðfðxÞÞ � xdðfðyÞÞ ¼ 0:

Applying Beidar et al. (1999, Theorem 1.2) it follows in particular that there exist p,q, p0, q0 2Q and l, l0 :A!C such that

fðxÞ ¼ dðxÞpþ xqþ lðxÞdðfðxÞÞ ¼ p0dðxÞ þ q0xþ l0ðxÞ

for all x2A. Applying d to the first identity and comparing the result obtained withthe second identity we get

d2ðxÞpþ dðxÞðdðpÞ þ qÞ þ xdðqÞ � p0dðxÞ � q0x 2 C:

Since char(A) 6¼ 2, note that Beidar et al. (1999, Theorem 1.2) can be applied again.Whence it follows that p¼ 0 and d(p)þ q¼ q2C, and so [f(x), x]¼ [xqþ l(x), x]¼ 0for every x2A, as desired.

Now suppose that A is a GPI ring. As already mentioned we may assume that dis defined on Q, and from Beidar et al. (1996, Theorem 4.5.3) it follows that thereis a2C such that b¼d(a) 6¼ 0. Let I be a nonzero ideal of A such that aI�A (Beidaret al., 1996, Proposition 2.2.3(ii)). For any y2 I we have

0 ¼ dð½fðayÞ; ay�Þ ¼ dða½fðayÞ; y�Þ ¼ b½fðayÞ; y� þ adð½fðayÞ; y�Þ¼ b½fðayÞ; y� þ adð½ay; fðyÞ�Þ ¼ b½fðayÞ; y� þ adða½y; fðyÞ�Þ¼ b½fðayÞ; y� þ ab½y; fðyÞ� ¼ b½fðayÞ � afðyÞ; y�:

Since 0 6¼ b2C, it follows from Bresar (1994, Corollary 4.2) that there exist l2C andm : I!C such that f(ay)� af(y)¼ lyþm(y) for all y2 I, and so we have

0 ¼ dð½fðayÞ; ay�Þ ¼ dð½afðyÞ þ lyþ mðyÞ; ay�Þ¼ dða2½fðyÞ; y�Þ ¼ dða2Þ½fðyÞ; y� ¼ 2ab½fðyÞ; y�

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Therefore [f(y), y]¼ 0 for all y2 I, and hence as noted above also for all y2A, asdesired.

Thus, we may assume from now on that d is an X-inner derivation, that is, thatthere exists a2QsnC such that d(x)¼ [a, x] for all x2A.

2.2. Reduction to the GPI Case

In this subsection we shall not only reduce the problem to the case when A is aGPI ring but also to the case when the element a is quite special. Here we shall relyheavily on results on functional identities with r-independent coefficients (Beidaret al., 2002, Secs. 2 and 3); we have to point out that the proof in this part is almostliterally the same as that of Beidar et al. (2002, Theorem 4.1). Nevertheless, we givedetails for the sake of completeness.

Note that (2) can now be written as

� afðxÞy� afðyÞxþ fðxÞyaþ fðyÞxa� yfðxÞa� xfðyÞaþ ayfðxÞ þ axfðyÞ ¼ 0: ð3Þ

Suppose that the elements 1 and a are 1-independent (see Beidar et al., (2002) for thedefinition). Then it follows from Beidar et al. (2002, Corollary 3.12) that f can beexpressed in three different ways:

fðxÞ ¼ xq1 þ axq2 þ l1ðxÞ þ l2ðxÞa; ð4Þ

fðxÞ ¼ xq3 þ axq4 þ l3ðxÞ þ l4ðxÞa; ð5Þ

fðxÞ ¼ q5x� q4xaþ l5ðxÞ � l2ðxÞa; ð6Þfor some qi2Q and li :A!C, i¼ 1, . . . , 5. Comparing (4) and (5) we see thatx(q1� q3)þ ax(q2� q4) always lies in the linear span of 1 and a. By Beidar et al.(2002, Corollary 3.5) we have q1¼ q3 and q2¼ q4. Accordingly, comparing (4) and(6) we obtain

q5x� q4xa� xq1 � axq4 ¼ l1ðxÞ � l5ðxÞ þ 2l2ðxÞa;

which by Beidar et al. (2002, Lemma 3.10) yields

q5x� q4xa� xq1 � axq4 ¼ l1ðxÞ � l5ðxÞ ¼ 2l2ðxÞ ¼ 0

for all x2A. Therefore l2¼ 0. Note that Beidar et al. (2002, Theorem 3.6)implies that q5¼ mþ na, �q4¼ gþ ta, q1¼mþ ga, q4¼ nþ ta for some m, n, g,t2C. However, gþ ta¼� n� ta gives g¼� n and t¼ 0. Using all these relationswe now see that (4) can be rewritten as

fðxÞ ¼ xðmþ gaÞ � axgþ l1ðxÞ ¼ mx� gdðxÞ þ l1ðxÞ:

If g 6¼ 0 then (1) implies d([d(x), x])¼ 0 for all x2A which e.g., by Bresar andVukman (1993, Lemma 1) yields d¼ 0, a contradiction.

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We may therefore assume that 1 and a are 1-dependent. According to Beidaret al. (2002, Proposition 2.9) this means that either A is algebraic of degree 2over C (equivalently, A is an order in B which is a 4-dimensional central simple alge-bra over C) or there exists b2C such that (a� b)Qs(a� b)¼C(a� b). The existenceof such element implies that A is a GPI ring, and hence its central closure B is a pri-mitive ring with nonzero socle and the associated division C-algebra is just 1-dimen-sional, i.e., it may be identified with C (see Beidar et al., 1996, Sec. 6.1). Moreover,a� b lies in soc(Qs), the socle of Qs, which in turn implies that a� b2B (Beidaret al., 1996, Theorem 4.3.6) and so also a2B.

2.3. Reduction to the Matrix Case

Since the central closure B of A is a vector space over its center C, there existsa C-subspace V of B such that B¼C�V. Let p :B!V be the canonical projectionof vector spaces.

Set g¼ pf :A!V. Clearly g is an additive map and g(x)� f(x)2C. Therefore[[g(x), x], a]¼ [[f(x), x], a]¼ 0 for all x2A, and hence

½½gðxÞ; y� þ ½gðyÞ; x�; a� ¼ 0 for all x; y 2 A: ð7Þ

Define h :B!B by

hXni¼1

gixi þ g

!¼Xni¼1

gigðxiÞ

for all x1, x2, . . . , xn2A, g, g1, g2, . . . , gn2C and for every positive integer n. We claimthat h is well-defined. So assume that

Pn

i¼1gixiþ g¼ 0 for some integer n and forsome xi2A, g, gi2C. Let y2A and note that by (7) we have

0 ¼Xni¼1

gið½gðxiÞ; y� þ ½gðyÞ; xi�Þ;a" #

¼Xni¼1

gigðxiÞ; y" #

þ gðyÞ;Xni¼1

gixi

" #; a

" #

¼Xni¼1

gigðxiÞ; y" #

þ ½gðyÞ;�g�; a" #

¼Xni¼1

gigðxiÞ; y" #

;a

" #:

That is, the composition of two derivations, x 7! [Pn

i¼1gig(xi), x] and x 7! [a, x], isequal to zero, and so using Posner’s theorem again we get

Pni¼1gig(xi)2C\V¼ 0.

Therefore h is well-defined. Clearly h is a C-linear map and h(x)¼ g(x) for allx2A. In particular, [[h(x), y]þ [h(y), x], a]¼ 0 for all x, y2A, and therefore, sinceh is C-linear, the latter identity in fact holds true for all x, y2B.

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Therefore we have reduced the proof to the case when A is a centrally closedprime GPI C-algebra with unity and f :A!A is a C-linear map. Moreover, denotingby C the algebraic closure of C and extending f to A�C C we see that there is no lossof generality in assuming that C is algebraically closed.

In the case when A(¼B¼Qs) is a 4-dimensional central simple algebra, weactually have AffiM2(C ) since C is algebraically closed. Let b be an eigenvalue ofa. Then a� b is a rank one matrix, so that (a� b)Qs(a� b)¼C(a�b) holds in thiscase as well.

We may replace the role of a by a� b (namely, d(x)¼ [a� b, x]), that is to say,we may take b¼ 0. Thus, without loss of generality we may assume that (in any case)a satisfies aQsa¼Ca 6¼ 0.

Up to this point we have followed the concept of the proof of Beidar et al. (2002,Theorem 4.1). From now on we use a different way.

Since a¼ aqa for some q2Qs, for all x2A we have

a½ fðxÞ; x� ¼ ½fðxÞ; x�a ¼ ½fðxÞ; x�aqa ¼ a½fðxÞ; x�qa 2 Ca: ð8Þ

As noted at the beginning of the proof it suffices to show that [f(s), s]¼ 0 for alls2 soc(A). Let s2 soc(A). Since a and [f(s), s] also lie in soc(A), by Litoff’s theorem(Beidar et al. 1996, Theorem 4.3.11) there is an idempotent e2 soc(A) such thata, s, [f(s), s]2 eAe; moreover, eAeffiMn(C) for some positive integer n. Definefe : eAe! eAe by fe(x)¼ ef(x)e and note that [[fe(x), x], a]¼ e[[f(x), x], a]e¼ 0 forall x2 eAe, and that [fe(s), s]¼ [f(s), s]. Thus, in order to prove that [f(s), s]¼ 0 itsuffices to show that [fe(x), x]¼ 0 for all x2 eAe.

We have thereby reduced the problem to the case when A is a matrix algebraMn(C) and a2A satisfies aAa¼Ca 6¼ 0, meaning that a is a rank one matrix.Without loss of generality we may assume that either a¼ e11 or a¼ e12.

2.4. The Matrix Case with a¼ e11

Let us first introduce some notation. If x2A¼Mn(C) is such that eijxekl¼ 0 forsome i, j, k, l2N¼f1, . . . ,ng, then clearly eujxekv¼ 0 for all u, v2N, and we shalldenote this also as e�jxek�¼ 0. This of course simply means that the (j, k)-entry ofx is 0. Similarly, eijx¼ 0 for some i, j2N means that the jth row of x is 0 andwe shall write this as e�jx¼ 0. The notation xek�¼ 0 has an analogous meaning.We identify elements from C by scalar matrices.

Our assumption is that [[f(x), x], e11]¼ 0 for all x2A. By (8) we have

e11½fðxÞ; x� ¼ ½fðxÞ; x�e11 2 Ce11

for all x2A. Linearizing we get

e11ð½fðxÞ; y� þ ½fðyÞ; x�Þ 2 Ce11 ð9Þ

ð½fðxÞ; y� þ ½fðyÞ; x�Þe11 2 Ce11 ð10Þ

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for all x, y2A. We have to show that [f(x), x]¼ 0 for all x2A, but actually we shallprove more, namely that there is l2C such that f(x)� lx2C for all x2A.

Pick i, j, k2N with i 6¼ 1, k 6¼ 1, j. Given any y2A it follows from (9) that

e11½fðeijÞ; y�ek1 ¼ e11ð½fðeijÞ; y� þ ½fðyÞ; eij�Þek1 2 Ce11ek1 ¼ 0;

that is, e11f(eij)yek1¼ e11yf(eij)ek1 for all y2A. A well-known result of Martindale,(1969, Theorem 2) implies that there is mij2C such that e11f(eij)¼ mije11 andf(eij)ek1¼ mijek1. That is,

e�1ðfðeijÞ � mijÞ ¼ 0 ¼ ðfðeijÞ � mijÞek� whenever i 6¼ 1 and k 6¼ 1; j. ð11Þ

Similarly, if j 6¼ 1 and k 6¼ 1, i we have by (10) that e1k[f(eij), y]e11¼ 0 which yieldsthat there is nij2C such that

ðfðeijÞ � nijÞe1� ¼ 0 ¼ e�kðfðeijÞ � nijÞ whenever j 6¼ 1 and k 6¼ 1; i. ð12Þ

Let i 6¼ 1 and j 6¼ 1. If n¼ 2, then it follows from e11[f(e22), e22]e222Ce11e22¼ 0 ande22[f(e22), e22]e112Ce22e11¼ 0 that e11f(e22)e22¼ 0¼ e22f(e22)e11. If n� 3, then wesee from (11) and (12) that mije11¼ e11f(eij)e11¼ nije11 and so mij¼ nij. Therefore,there is lij2C such that

fðeijÞ ¼ lijeij þ mij whenever i 6¼ 1 and j 6¼ 1. ð13Þ

Next we consider the case when j¼ 1 and i 6¼ 1. By (11), e�1(f(ei1)� mi1)¼ 0¼(f(ei1)� mi1)ek� for all k 6¼ 1. Moreover, if r 6¼ 1, i then it follows from (10) thate1r([f(ei1), err]þ [f(err), ei1])e112Ce1re11¼ 0 and from (13) that [f(err), ei1]¼ 0, andso e�r(f(ei1)� mi1)e1�¼ 0. Therefore there is li12C such that

fðei1Þ ¼ li1ei1 þ mi1 whenever i 6¼ 1. ð14Þ

Similarly, if i¼ 1 and j 6¼ 1, then it follows from (12) that (f(e1j)� n1j) e1�¼0¼ e�k(f(e1j)� n1j) for all k 6¼ 1, and from (9) and (13) that e�1(f(e1j)� n1j) er�¼ 0for all r 6¼ 1, j. Therefore, there exists l1j2C such that

fðe1jÞ ¼ l1je1j þ n1j whenever j 6¼ 1. ð15Þ

Let i 6¼ 1 and j 6¼ 1. From (9), (13) and (15) it follows that e11([f(e1i), eij]þ[f(eij), e1i])¼ (l1i� lij)e1j2Ce11 and so l1i¼ lij, and from (10), (13) and (14) wesee that ([f(ej1), eij]þ [f(eij), ej1])e11¼ (lij� lj1)ei12Ce11 and so lij¼ lj1. That is,lij’s are the same for all i 6¼ 1 or j 6¼ 1, and we denote this element by l.

Let k 6¼ 1. We see from e�1[f(e11), e11]ek�¼ 0¼ e�k[f(e11), e11]e1� that

e�1fðe11Þek� ¼ 0 ¼ e�kfðe11Þe1�

and from

0 ¼ e11ð½fðe11Þ; e1k� þ ½fðe1kÞ; e11�Þek1 ¼ e11fðe11Þe11 � e1kfðe11Þek1 � le11

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that

e1kfðe11Þek1 ¼ e11fðe11Þe11 � le11:

Moreover, taking j 6¼ 1 and r 6¼ 1, j, we have

e�1ð½fðe11Þ; e1j� þ ½fðe1jÞ; e11�Þer� ¼ 0

and, by (15), [f(e1j), e11]er�¼ 0, so that e�jf(e11)er�¼ 0. Therefore,

fðe11Þ � le11 2 C: ð16Þ

By (13–16) we see that f(x)� lx2C holds true for matrix units x¼ eij, but then,by the linearity of f, it holds true for all x2A. The proof is thus complete.

2.5. The Matrix Case with a¼ e12

The final case that has to be considered is when a linear map f on A¼Mn(C)satisfies [[f(x), x], e12]¼ 0 for all x2A. Again we shall prove that there is l2C suchthat f(eij)� leij2C for all matrix units eij. This proof, however, is longer than theone of the previous case.

By assumption we have e11[[f(x), x], e12]e22¼ 0 for all x2A, that is

e11½fðxÞ; x�e12 ¼ e12½fðxÞ; x�e22: ð17Þ

According to (8), we have e12[f(x), x]¼ [f(x), x]e122Ce12. Therefore, for all x, y2Awe have

e12ð½fðxÞ; y� þ ½fðyÞ; x�Þek2 2 Ce12ek2 ¼ 0 whenever k 6¼ 2; ð18Þe1kð½fðxÞ; y� þ ½fðyÞ; x�Þe12 2 Ce1ke12 ¼ 0 whenever k 6¼ 1: ð19Þ

Pick i, j, k2N with j 6¼ 1, k 6¼ 1, i. Given any y2A it follows from (19) that

e1k½fðeijÞ; y�e12 ¼ e1kð½fðeijÞ; y� þ ½fðyÞ; eij�Þe12 ¼ 0;

that is, e1kf(eij)ye12¼ e1kyf(eij)e12 for all y2A. By Martindale (1969, Theorem 2)there exists mij2C such that e1kf(eij)¼ mije1k and f(eij)e12¼ mije12. That is,

ðfðeijÞ � mijÞe1� ¼ 0 ¼ e�kðfðeijÞ � mijÞ whenever j 6¼ 1 and k 6¼ 1; i. ð20ÞSimilarly, there exists nij2C such that

e�2ðfðeijÞ � nijÞ ¼ 0 ¼ ðfðeijÞ � nijÞek� whenever i 6¼ 2 and k 6¼ 2; j. ð21ÞIf i 6¼ 2 and j 6¼ 1, then we see from (20) and (21) that

ðfðeijÞ � mijÞe12 ¼ 0 ¼ ðfðeijÞ � nijÞe12;hence mije12¼ f(eij)e12¼ nije12, and so mij¼ nij. Therefore,

fðeijÞ � lijeij � mij 2 Ce12 þ Ce1j þ Cei2 whenever i 6¼ 2 and j 6¼ 1 ð22Þ

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for some lij2C. In particular, there is l122C such that

fðe12Þ ¼ l12e12 þ m12: ð23Þ

Next we consider the case when i¼ 2 and j 6¼ 1. Since e12[f(e22), e22]e122Ce12e12¼ 0we have e12f(e22)e12¼ 0 which together with (20) shows that there is m2j2C such that

ðfðe2jÞ � m2jÞe1� ¼ 0 ¼ e�kðfðe2jÞ � m2jÞ for all k 6¼ 1; 2: ð24Þ

Let r 6¼ 1, 2, j. By (18) and (22) we have

0 ¼ e�2ð½fðe2jÞ; err� þ ½fðerrÞ; e2j�Þer�¼ e�2fðe2jÞer�: ð25Þ

Further, if j 6¼ 1, 2 then, by (18), e12[f(e2j)� m2j, e2j]ej2¼ 0 and using (24) we see that

e12ðfðe2jÞ � m2jÞe22 ¼ 0: ð26Þ

Similarly, if j¼ 1 and i 6¼ 2, it follows from e12f(e11)e12¼ 0, (since e12[f(e11), e11]e122Ce12e12¼ 0) and (21) that there is ni12C such that

e�2ðfðei1Þ � ni1Þ ¼ 0 ¼ ðfðei1Þ � ni1Þek� for all k 6¼ 1; 2: ð27Þ

Taking r 6¼ 1, 2, i, (19) shows that e�r([f(ei1), err]þ [f(err), ei1])e1�¼ 0, and (22) showsthat

e�rfðei1Þe1� ¼ 0: ð28ÞFurther, given i 6¼ 1, 2, (19) gives e�i[f(ei1)� ni1, ei1]e1�¼ 0 and (27) yields

e�1ðfðei1Þ � ni1Þe1� ¼ 0: ð29ÞLet k 6¼ 1, 2. By (18) and (19) we have

e�2½fðe21Þ; e21�ek� ¼ 0 ¼ e�k½ fðe21Þ; e21�e1�;that is

e�1fðe21Þek� ¼ 0 ¼ e�kfðe21Þe2�; ð30Þand

e�2ð½fðe21Þ; e2k� þ ½fðe2kÞ; e21�Þe1� ¼ 0:

Therefore, using (24) and (26) we see that

e�kfðe21Þe1� ¼ 0; ð31Þand

e�2ð½fðe21Þ; ek1� þ ½ fðek1Þ; e21�Þe1� ¼ 0;

and so, using (27) and (29),

e�2fðe21Þek� ¼ 0: ð32Þ

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Further, by (18) we have

0 ¼ e12½fðe21Þ; e21�e12 ¼ e12fðe21Þe22 � e11fðe21Þe12; ð33Þ

and by (17) we have

0 ¼ e11½fðe21Þ; e21�e12 � e12½fðe21Þ; e21�e22 ¼ 2e11fðe21Þe22: ð34Þ

Let j 6¼ 1. Again using (17) we see that

e11ð½fðe21Þ; e2j� þ ½fðe2jÞ; e21�Þe12 ¼ e12ð½fðe21Þ; e2j� þ ½fðe2jÞ; e21�Þe22:

Since e12[f(e21), e2j]e22¼ 0 (if j 6¼ 2 by (30)), it follows that e11f(e2j)e22¼� e11f(e2j)e22, that is,

e�1fðe2jÞe2� ¼ 0 whenever j 6¼ 1: ð35Þ

Similarly, let i 6¼ 2. From (17) it follows that

e12ð½fðe21Þ; ei1� þ ½fðei1Þ; e21�Þe22 ¼ e11ð½fðe21Þ; ei1� þ ½fðei1Þ; e21�Þe12;

which in view of e11[f(e21), ei1]e12¼ 0 (if i 6¼ 1 by (30)) yields �e11f(ei1)e22¼e11f(ei1)e22, that is,

e�1fðei1Þe2� ¼ 0 whenever i 6¼ 2: ð36Þ

Let i 6¼ 2, j 6¼ 1 and (i, j) 6¼ (1, 2). By (17) we have

e11ð½fðeijÞ; e21� þ ½fðe21Þ; eij�Þe12 ¼ e12ð½fðeijÞ; e21� þ ½fðe21Þ; eij�Þe22;

hence, using (31) (in the case i¼ 1, j 6¼ 2) and (32) (in the case j¼ 2, i 6¼ 1), we havee11f(eij)e22¼� e11f(eij)e22 and so

e�1fðeijÞe2� ¼ 0 whenever i 6¼ 2; j 6¼ 1 and ði; jÞ 6¼ ð1; 2Þ. ð37Þ

Further, if i, j 6¼ 1, 2, then it follows from (18) and (19) that

e�ið½fðeijÞ; e21� þ ½fðe21Þ; eij�Þe1� ¼ 0 ¼ e�2ð½fðeijÞ; e21� þ ½fðe21Þ; eij�Þej�

and so, by (31) and (32), we have

e�ifðeijÞe2� ¼ e�jfðe21Þe1� ¼ 0 and

e�1fðeijÞej� ¼ e�2fðe21Þei� ¼ 0 whenever i; j 6¼ 1; 2:ð38Þ

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Note that from (22), (37) and (38) it follows that

fðeijÞ ¼ lijeij þ mij whenever i 6¼ 2; j 6¼ 1 and ði; jÞ 6¼ ð1; 2Þ: ð39Þ

Let k 6¼ 1, 2. By (17) we see that

e11ð½fðe22Þ; ek1� þ ½fðek1Þ; e22�Þe12 ¼ e12ð½fðe22Þ; ek1� þ ½ fðek1Þ; e22�Þe22

and

e11ð½fðe11Þ; e2k� þ ½fðe2kÞ; e11�Þe12 ¼ e12ð½fðe11Þ; e2k� þ ½fðe2kÞ; e11�Þe22;

which in turn implies

e�1fðe22Þek� ¼ 0 and

e�kfðe11Þe2� ¼ 0 whenever k 6¼ 1; 2:ð40Þ

Therefore, it follows from (24), (25), (35), (40) and (27), (28), (36), (40) that thereexist l22, l112C such that

fðe22Þ ¼ l22e22 þ m22 and

fðe11Þ ¼ l11e11 þ m11:ð41Þ

Let i, j 6¼ 1, 2. By (18) we have

e12ð½fðe2jÞ; e21� þ ½fðe21Þ; e2j�Þei2 ¼ 0;

that is

�e11fðe2jÞei2 ¼ e1jfðe21Þei2; if i 6¼ j; and

�e11fðe2jÞei2 ¼ e1jfðe21Þei2 � e12fðe21Þe22; if i ¼ j:ð42Þ

Similarly, by (19) we have

e1jð½fðei1Þ; e21� þ ½fðe21Þ; ei1�Þe12 ¼ 0;

that is

e1jfðei1Þe22 ¼ �e1jfðe21Þei2; if i 6¼ j; and

e1jfðei1Þe22 ¼ �e1jfðe21Þei2 þ e11fðe21Þe12; if i ¼ j:ð43Þ

Comparing (42) and (43) (and (33) in case i¼ j), we have

e11fðe2jÞei2 ¼ e1jfðei1Þe22: ð44Þ

Further, it follows from (17) that

e11ð½fðei1Þ; e2j� þ ½fðe2jÞ; ei1�Þe12 ¼ e12ð½fðei1Þ; e2j� þ ½fðe2jÞ; ei1�Þe22;

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hence

e11fðe2jÞei2 ¼ �e1jfðei1Þe22: ð45Þ

By (44) and (45) we have

e�1fðe2jÞei� ¼ 0 ¼ e�jfðei1Þe2� whenever i; j 6¼ 1; 2: ð46Þ

Therefore, it follows from (24), (25), (26), (35), (46) and (27), (28), (29), (36), (46)that there exist l2j, lj12C such that

fðe2jÞ ¼ l2je2j þ m2j and

fðej1Þ ¼ lj1ej1 þ nj1 whenever j 6¼ 1; 2ð47Þ

and so, by (42), we have

e�jfðe21Þei� ¼ 0 whenever i 6¼ j; i; j 6¼ 1; 2 and

e1ifðe21Þei2 ¼ e12fðe21Þe22 whenever i 6¼ 1; 2:ð48Þ

Hence, we see from (30), (31), (32), (33), (34), (48) that there exist l212C such that

fðe21Þ ¼ l21e21 þ m21: ð49Þ

Therefore, from (23), (39), (41), (47), (49) it follows that there exist lij2C such that

fðeijÞ ¼ lijeij þ mij for all i; j.

It remains to show that all lij’s are the same. Let j 6¼ 2. Since by (18)

e12ð½fðe2kÞ; ekj� þ ½fðekjÞ; e2k�Þej2 ¼ 0;

we have (l2k� lkj)e12¼ 0 and so l2k¼ lkj for all k. Similarly, for i 6¼ 1, by (19)we have

e1ið½fðeikÞ; ek1� þ ½fðek1Þ; eik�Þe12 ¼ 0;

hence (lik� lk1)e12¼ 0 and so lik¼ lk1 for all k. Therefore

lij ¼ l for all i; j with ði; jÞ 6¼ ð1; 2Þ:

Since

0 ¼ ½½fðe12Þ; e21� þ ½fðe21Þ; e12�; e12�¼ ðl12 � lÞ½e11 � e22; e12�¼ 2ðl12 � lÞe12

we have l12¼ l, as desired.The proof of Theorem 1 is thus complete.

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ACKNOWLEDGMENT

I am deeply thankful to Professor M. Bres�ar for initiating this problem and formany helpful suggestions. This article would have never been written without hishelp and encouragement. I also thank the referee for careful reading and noticingthat the first version of the article contained an incompleteness in the last part ofthe proof.

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Received August 2002Revised October 2002

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